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Basis and DimensionIn the preceding section, we observed that
e_{1} = (1,0,0,...,0)
e_{2} = (0,1,0,...,0)
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e_{n} = (0,0,0,...,1)
is a set of generators for the vector space F^{n}. Thus, every element v of F^{n} can be written in the form (1)
v = _{i}e_{i}, _{i} F.
Actually, the set S of generators has a very special property: The elements _{i} in (1) are uniquely determined. Indeed,
_{i}e_{i} = (_{1}, _{2}, ..., _{n}),
so that if (1) holds, and Definition 1: Let V be a vector space over F. A basis of V is a subset {e_{i}} of V (finite or infinite) with the property that every element v of V can be uniquely written in the form
v = _{i}e_{i}.
(Here we always assume that if {e_{i}} is infinite, then Example 1: {e_{i}, ..., e_{n}} is a basis for F^{n}. Example 2: {e_{1}  e_{2},e_{1} + e_{2}} is a basis for F^{2} since
v = (v_{1}  v_{2})/2(e_{1}  e_{2}) + (v_{1} + v_{2})/2(e_{1} + e_{2}),
and if The main goal of this section is to show that any two bases of a vector space V have the same number of elements, in the following sense: If one basis of V has a finite number of elements, then every basis has a finite number of elements, and every basis has the same number. If one basis of V is infinite, then every basis is infinite. Assuming this result for the moment, let us make the following definition.
Definition 2: Let V be a vector space. If V has a finite set of generators, then we say that V is finitedimensional and we define its dimension, denoted For example, let Assume no that
v = v_{1} + ... + _{n}v_{n} (_{i} F).
We assert that this representation is unique. For if
v_{1} = _{2}v_{2} + ... + _{n}v_{n},
where
w = _{1}v_{1} + ... + _{n}v_{n}, (_{i} F)
= (_{2} + _{1}_{2})v_{2} + ... + (_{n} + _{1}_{n})v_{n},
so that w can be written as a linear combination of v_{2}, ..., v_{n}. Then, since
Definition 3: Let
_{1}s_{1} + _{2}s_{2} + ...+ _{n}s_{n} = 0.
If S is not linearly dependent, then we say that S is linearly independent. Example 3: Let
(2) · e_{1} + 1 · (2e_{1}) = 0.
Example 4: Let V be arbitrary and let Example 5: Let V be arbitrary and let S be a basis of V. Then S is linearly independent, since if
_{1}s_{1} + _{2}s_{2} + ... + _{n}s_{n} = 0, _{i} F, s_{i} S,
then since
0 = 0 · s_{1} + 0 · s_{2} + ... + 0 · s_{n},
the fact that 0 is uniquely representable in terms of the basis implies that _{1} = _{2} = ... = _{n} = 0.
Proposition 4: Let V be a vector space and let Proof: If S is a basis, then S is clearly a set of generators for V. Moreover, S is linearly independent by Example 5 above.
Let
v = _{s}s,
where all but a finite number of the _{s} are 0. Let us show that the representation (2) is unique. Indeed if
v = _{s}s, _{s} F,
then
0 = v  v = (_{s}  _{s})s.
But since S is linearly independent, we see that Theorem 5: Let V be a vector space. Then any two bases of V contain the same number of elements. Proof: Let us consider the special case where V has a finite basis {e_{1}, ..., e_{n}}. We will prove that if {f_{i}}_{iI} is any other basis of V, then it is finite and contains n elements. Consideration of the special case suffices, for then if V has one infinite basis, then the special case implies that all bases must be infinite. Since {f_{i}}_{iI} is linearly independent, all the f_{i} are nonzero. Let f_{1} be one of the f_{i}. Since {e_{1}, ..., e_{n}} is a basis for V, there exist
f_{1} = e_{1} + ... + _{n}e_{n}.
Since f_{1} 0, not all _{j} are zero. Let us renumber the e_{j} so that
e_{1} = _{1}^{1}f_{1}  _{1}^{1}_{2}e_{2}  ...  _{1}^{1}_{n}e_{n}.
v = _{1}e_{1} + ... + _{n}e_{n}.
Therefore, by (3),
v = _{1}'f_{1} + _{2}'e_{2} + ... + _{n}'f_{n}.
Thus, {f_{1},e_{2},...,e_{n}} is a set of generators for V. If {f_{1},e_{2},...,e_{n}} is linearly dependent, then there exist
_{1}f_{1} + _{2}e_{2} + ... + _{n}e_{n} = 0,
where _{1},...,_{n} are not all 0. If
_{1}e_{1} + (_{2} + _{2})e_{2} + ... + (_{n} + _{n})e_{n} = 0, _{1} 0,
which contradicts the fact that {e_{1},...,e_{n}} are linearly independent. The contradiction proves that {f_{1},e_{2}...,e_{n}} is linearly independent, so that by Proposition 4, {f_{1},e_{2}...,e_{n}} is a basis of V. If
f_{2} = _{1}f_{1} + _{2}e_{2} + ... + _{n}e_{n},
_{1},...,_{n} not all 0. If Remarks: 1. It is possible to strengthen Theorem 5 in case V has infinite dimension. Namely, it is possible to show that not only are any two bases of V is infinite, but given any two bases of V, there exists a bijection of one of these into the other. 2. It is not clear from what we have said that any vector space has a basis. In order to prove this assertion, it is necessary to appeal to Zorn's lemma. If V is finitedimensional, then the existence of a basis of V was proved. In most of what follows, we will be concerned only with finitedimensional vector spaces, so we will not give proof of the existence of a basis in the most general case. 
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